Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $q = \dfrac{k^2 + 15k + 50}{k + 5} \div \dfrac{-9k^2 - 90k}{-6k - 36} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{k^2 + 15k + 50}{k + 5} \times \dfrac{-6k - 36}{-9k^2 - 90k} $ First factor the quadratic. $q = \dfrac{(k + 10)(k + 5)}{k + 5} \times \dfrac{-6k - 36}{-9k^2 - 90k} $ Then factor out any other terms. $q = \dfrac{(k + 10)(k + 5)}{k + 5} \times \dfrac{-6(k + 6)}{-9k(k + 10)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (k + 10)(k + 5) \times -6(k + 6) } { (k + 5) \times -9k(k + 10) } $ $q = \dfrac{ -6(k + 10)(k + 5)(k + 6)}{ -9k(k + 5)(k + 10)} $ Notice that $(k + 5)$ and $(k + 10)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -6\cancel{(k + 10)}(k + 5)(k + 6)}{ -9k(k + 5)\cancel{(k + 10)}} $ We are dividing by $k + 10$ , so $k + 10 \neq 0$ Therefore, $k \neq -10$ $q = \dfrac{ -6\cancel{(k + 10)}\cancel{(k + 5)}(k + 6)}{ -9k\cancel{(k + 5)}\cancel{(k + 10)}} $ We are dividing by $k + 5$ , so $k + 5 \neq 0$ Therefore, $k \neq -5$ $q = \dfrac{-6(k + 6)}{-9k} $ $q = \dfrac{2(k + 6)}{3k} ; \space k \neq -10 ; \space k \neq -5 $